Worked Solutions
Statistical Analysis — Worked Solutions (HSC Maths Extension 1)
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Worked examples for HSC Maths Extension 1 the binomial distribution. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — A binomial probability
Question
A fair six-sided die is rolled $5$ times. Find the probability of obtaining exactly two sixes. Give your answer as an exact fraction.
Solution
Binomial with $n = 5$, $p = \tfrac16$ (a six), $q = \tfrac56$. Want exactly $2$ successes:
$$P(X = 2) = \binom{5}{2}\left(\frac16\right)^2\left(\frac56\right)^3.$$
$\binom{5}{2} = 10$, $\left(\tfrac16\right)^2 = \tfrac{1}{36}$, $\left(\tfrac56\right)^3 = \tfrac{125}{216}$.
$$P(X=2) = 10\cdot\frac{1}{36}\cdot\frac{125}{216} = \frac{1250}{7776} = \frac{625}{3888}.$$
State $n$, $p$ and the number of successes up front — that's the setup mark.
Each roll is an independent trial that either gives a six (success, $p = \tfrac16$) or not (failure, $q = \tfrac56$), with a fixed number of rolls — that's exactly a binomial situation. The formula is $P(X = k) = \dbinom{n}{k}p^k q^{n-k}$.
Here $n = 5$ and we want $k = 2$ sixes:
$$P(X = 2) = \binom{5}{2}\left(\frac16\right)^2\left(\frac56\right)^3.$$
Work through the pieces: $\binom{5}{2} = 10$ counts which two of the five rolls are sixes; $\left(\tfrac16\right)^2 = \tfrac{1}{36}$ is the chance those two are sixes; $\left(\tfrac56\right)^3 = \tfrac{125}{216}$ is the chance the other three are not.
$$10 \cdot \frac{1}{36} \cdot \frac{125}{216} = \frac{1250}{7776} = \frac{625}{3888}.$$
The $\binom{5}{2}$ matters because the two sixes could land on any pair of rolls — we count all those ways.
- Binomial: $n = 5$, $p = \frac16$, $q = \frac56$, $k = 2$
- $P(X=2) = \binom{5}{2}\left(\frac16\right)^2\left(\frac56\right)^3$
- $= 10\cdot\frac{1}{36}\cdot\frac{125}{216} = \frac{1250}{7776} = \frac{625}{3888}$
Where the marks go
- 1 mark: Identifies binomial with $n = 5$, $p = \frac16$
- 1 mark: Correct expression $\binom{5}{2}(\frac16)^2(\frac56)^3$
- 1 mark: Exact answer $\frac{625}{3888}$
Key idea
For "exactly $k$ successes" use $P(X=k) = \binom{n}{k}p^k q^{n-k}$, with the binomial coefficient counting which trials succeed.
Example 2 — Mean and variance of a binomial
Question
A multiple-choice quiz has $80$ questions, each with $4$ options. A student guesses every answer. Let $X$ be the number of correct answers. Find the mean and standard deviation of $X$.
Solution
Binomial: $n = 80$, $p = \tfrac14$, $q = \tfrac34$.
Mean: $E(X) = np = 80 \cdot \tfrac14 = 20$.
Variance: $\operatorname{Var}(X) = npq = 80 \cdot \tfrac14 \cdot \tfrac34 = 15$.
Standard deviation: $\sqrt{15} \approx 3.87$.
Use $np$ for the mean and $npq$ for the variance — then square-root for the SD. Don't quote the variance when asked for SD.
Guessing 80 independent questions, each correct with probability $p = \tfrac14$, is a binomial experiment with $n = 80$. For a binomial there are tidy formulas so we don't sum anything by hand.
The mean (expected number correct) is $E(X) = np$: on average a guesser gets $80 \cdot \tfrac14 = 20$ right. That makes sense — a quarter of 80.
The variance is $\operatorname{Var}(X) = npq$, where $q = 1 - p = \tfrac34$: $80 \cdot \tfrac14 \cdot \tfrac34 = 15$.
The question asks for the standard deviation, which is the square root of the variance: $\sqrt{15} \approx 3.87$.
We square-root because standard deviation is measured in the same units as $X$ (questions), whereas variance is in squared units.
- $n = 80$, $p = \frac14$, $q = \frac34$
- Mean $= np = 20$
- Variance $= npq = 80\cdot\frac14\cdot\frac34 = 15$
- SD $= \sqrt{15} \approx 3.87$
Where the marks go
- 1 mark: Identifies binomial with $n = 80$, $p = \frac14$
- 1 mark: Mean $E(X) = np = 20$
- 1 mark: Variance $npq = 15$
- 1 mark: Standard deviation $\sqrt{15} \approx 3.87$
Key idea
For a binomial, mean $= np$ and variance $= npq$; the standard deviation is $\sqrt{npq}$.